Answer
$${y}=c_{1}\exp^{-\frac{1}{3}x}+c_{2}\exp^{\frac{1}{2}x}$$
Work Step by Step
$$6y''-y'-y=0$$Let $y-e^{\lambda{x}}$ so that $(\ln{y})'=\lambda$.
$$6{\lambda}^2-{\lambda}-1=0$$
$$(3\lambda+1)(2\lambda-1)=0$$
$$\lambda_{1,2}=-\frac{1}{3},\frac{1}{2}$$
$$\therefore{y}=c_{1}\exp^{-\frac{1}{3}x}+c_{2}\exp^{\frac{1}{2}x}$$